Find the volume of the solid of revolution when the region bounded by the curve y = ¼ (4 – x)² and y = ½ (4 – x²) and the y-axis when the bounded region R rotates 2pie rad about the y-axis. I don't even have a single idea on solving it. Its my math assignment and I'm desperate for help

I suggest you first sketch out the two curves, so you get an idea what the cross-section of the shape looks like. Your problem statement left out the range of integration - these curves don't intersect so they should have specified the range of x to use.

If you were asked to simply find the area between the two curves I assume you would know how to do that - you would have an integral of the difference between the two curves, evaluated from x = 0 to x = whatever the upper limit of x is, correct? That area is the sum of all the infinitesimal little rectangles, each of which has a height equal to the difference between the two curves and whose width is dx. Well, if you imagine the shape being rotated about the y-axis you end up with a shape whose volume is the sum of infinitesimal cylinders, each of which is centered on the y-axis and has a volume equal to its circumference of 2 pi x, times its height (which again is equal to the difference between the curves), times its width dx. Try it, and post back with your attempt.